To multiply the rational expressions \(\frac{x^2 + 2x - 3}{x^2 + 3x - 4} \cdot \frac{x^2 + 6x + 8}{x^2 - 9}\) and simplify, let's start by factoring each polynomial.
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Factor \(x^2 + 2x - 3\):
This factors as \((x + 3)(x - 1)\). -
Factor \(x^2 + 3x - 4\): This factors as \((x + 4)(x - 1)\).
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Factor \(x^2 + 6x + 8\): This factors as \((x + 2)(x + 4)\).
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Factor \(x^2 - 9\): This is a difference of squares and factors as \((x + 3)(x - 3)\).
Now we can rewrite the expression with these factored forms:
\[ \frac{(x + 3)(x - 1)}{(x + 4)(x - 1)} \cdot \frac{(x + 2)(x + 4)}{(x + 3)(x - 3)} \]
Next, we multiply the numerators and the denominators:
Numerator:
\[ (x + 3)(x - 1)(x + 2)(x + 4) \]
Denominator:
\[ (x + 4)(x - 1)(x + 3)(x - 3) \]
Now we combine the fractions:
\[ \frac{(x + 3)(x - 1)(x + 2)(x + 4)}{(x + 4)(x - 1)(x + 3)(x - 3)} \]
Now we can cancel out the common factors in the numerator and the denominator:
- \((x + 3)\)
- \((x - 1)\)
- \((x + 4)\)
This leaves us with:
\[ \frac{(x + 2)}{(x - 3)} \]
Thus, the product in simplest form is:
\[ \frac{x + 2}{x - 3} \]
So, the final answer is:
\(\frac{x + 2}{x - 3}\)